# Image of open set under unbounded linear functional

Let $$X$$ a normed space over a field $$\mathbb{K}=\mathbb{R}$$ or $$\mathbb{C}$$ and let $$f$$ a unbounded linear functional. Let $$U \neq \emptyset$$ a open set in $$X$$. I have to prove that $$f(U)=\mathbb{K}$$.

My attempt. Let $$x \in \mathbb{K}$$ and since $$U \neq \emptyset$$ exists $$u \in U$$ and since $$U$$ is open exists $$\varepsilon>0$$ such that $$B(u, \varepsilon) \subseteq U$$. Since $$f$$ is unbounded exists $$y \in X$$ such that $$\vert f(x) \lvert > \varepsilon \vert \vert y \vert \vert$$. I have noticed that $$f \left( \dfrac{x}{f(y)} y \right)=\dfrac{x}{f(y)} f(y)=x.$$ And also: $$f \left( \dfrac{x}{f(u)} u \right)=\dfrac{x}{f(u)} f(u)=x.$$ I would need to try that $$\dfrac{x}{f(y)} y$$ or $$\dfrac{x}{f(u)} u$$ is in $$U$$. And to do so, prove that is in $$B(u, \varepsilon)$$. But I'm stuck on it.

• It suffices to show that the image of the open unit ball $B(0,1)$ is equal $K.$ Indeed any open set contains a ball $B(x_0,r).$ Then $f(B(x_0,r))= f(x_0)+rfB(0,1)).$ By discontinuity the image of the unit ball contains numbers in $K$ of arbitrary large absolute value. Then using $f(\lambda x)=\lambda f(x)$ we can get the conclusion. May 14 at 1:55